Find the point closest to the line sqroot(X+1) from the point (3,0).
d = [(x - 3) + (y - 0)]^1/2
d = [(x - 3) + (y)]1/2
Do I now substitute in the equation y = sqroot(X+1) and solve?
Yes, and take the derivative and set to zero for the min distance.
You forgot the squares:
d = [(x - 3)^2 + y^2]1/2
It is easier to minimize d^2:
d^2 = (x-3)^2 + y^2
Insert in here y^2 = x+1 and set the derivative of d^2 w.r.t. x equal to zero.
Thanks both of you.
To find the point closest to the line sqroot(X+1) from the point (3,0), you can use the distance formula. Here are the steps:
1. Start with the distance formula: d = [(x - x₁)² + (y - y₁)²]¹/², where (x₁, y₁) represents the coordinates of the point (3,0).
2. Substitute the value of x₁ as 3 and y₁ as 0: d = [(x - 3)² + (y - 0)²]¹/².
3. Now, you need to substitute the equation of the line sqroot(X+1) for y in the distance formula.
So, y = sqroot(X+1).
4. Replace y in the distance formula with sqroot(X+1): d = [(x - 3)² + (sqroot(X+1) - 0)²]¹/².
5. Simplify the equation inside the square brackets: d = [(x - 3)² + (sqroot(X+1))²]¹/².
6. Since you are looking for the point closest to the line, you want to minimize the distance d. To find the minimum, you can take the derivative and set it equal to zero.
Calculating the derivative and setting it equal to zero will give you the critical points and allow you to determine the x-coordinate of the point closest to the line.
7. Solve for x by taking the derivative of d with respect to x and setting it equal to zero.
d'x = 0 gives you the critical points.
8. Once you have the x-coordinate of the point closest to the line, substitute it back into the equation y = sqroot(X+1) to find the corresponding y-coordinate.
By following these steps, you should be able to find the point closest to the line sqroot(X+1) from the point (3,0).